Note: The associated data bits for the parity bit are chosen by the parity bit's exponentiation. F.e. P1 = 2^0 = 1, so you check & skip bits always with 1 step. CHECK, SKIP, CHECK, SKIP etc. All your check-bits are the ones that will be used for parity later. F.e. P2 = 2^1 = 2, this would result in CHECK, CHECK, SKIP, SKIP, CHECK, CHECK, SKIP, SKIP etc. F.e. P3 = 2^2 = 4, this would result in 4 times CHECK, 4 times SKIP, 4 times CHECK, 4 times SKIP etc. And the bits that are checked always have to be data bits. Try it with the 7 bit array in the upper right corner of the video. You will see how the associations P1 -> D3,D5,D7 P2 -> D3D6D7 P4-> D5D6D7 are easily made.
Other way is to check and group similar bit group positions . P1 index is 1 -> 001 Check of the bits with 1 at that position and group them .. Similarly P2 is 010 Group again..
We don't have to cram anything out. p1 takes care of data bits which have 1 in rightmost place(i.e lsb) and p2 takes care of data bits which have 1 at second rightmost place after lsb and so on .Also for every r>=2,we have 2^r-1 length code block and length of message data is 2^r-r-1.
Thanks sir!! It's really helpful.. I think our hardware teacher must watch this video before giving that boring lecture to us !! anyway free of worry after watching your video!! Thanks Once again sir!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
The Hamming Code is an error-correcting code used to detect and correct single-bit errors in digital data transmission. To understand the process of using the Hamming Code for error detection and decision-making, let's break it down into stages: 1. Encoding: In this stage, the original data (message) is encoded using the Hamming Code. The encoding process adds extra parity bits (check bits) to the original data. These parity bits are strategically placed to help detect and correct single-bit errors. The total number of bits in the encoded message will be greater than the original data. 2. Transmission: After encoding, the message is transmitted through a communication channel. During transmission, there is a possibility of errors, such as flipping a single bit, due to noise or other factors. 3. Receiving: The encoded message is received at the destination. However, due to the transmission errors, the received message may have one or more bits in error. 4. Error Detection: In this stage, the received message is checked for errors using the parity bits added during encoding. The receiver calculates the parity of the received message and compares it with the parity bits. If the parity bits match the calculated parity, it indicates that no errors occurred during transmission. If the parity bits do not match, it means that there is at least one error in the received message. 5. Error Correction: If an error is detected in the previous step, the receiver uses the information from the parity bits to identify the location of the error and correct it. The Hamming Code allows for the correction of single-bit errors. 6. Decision-making: After error correction (if applicable), the receiver decides whether the message is error-free and ready for use or if it needs further action, such as retransmission of the message. In summary, the stages involved in error detection and decision-making using the Hamming Code include encoding, transmission, receiving, error detection, error correction, and decision-making based on the corrected or uncorrected message.
hey you are so gooddddd at teaching .Thanks for saving my mid term test... Imagine spending 5 weeks listening to lecture but you cant learn shit.But a 12 minutes video make you undesratnd whole concept :D
+Andrew Morris the answer actually lies in the binary value of these positions lets say the value of position one is 1 in decimal and 001 in binary (considering that 7 can be represented in 3 bits), now Parity P1 lies on 001, we will select those values for which the 1 lies on this position, i-e the rightmost position (it may have more ones but 1 should be presented on the indicated position of the parity bit), naturally we will find these are D3 (011),D5(101), D7(111) so these three have one at the right most position. That is why we select them. Also the parity bits will have only Single 1 in them so issue of multiple ones will not be there, Practice it Hope it answers the question.
Gayatri Kulkarni First question is when will P5 needed? When a longer bits integer is used right? so then longer or bigger number will be represented in more binary digits i-e more then 3 bits (000,010 etc) so then binary digits will increase to the minimum no of binary digits to represent that number and that will how you will represent P5
Let a data bit sequence M=1110100001 is transmitted but the receiver receives the sequence with any one bit corrupted. Use hamming code to identify the corrupted bit position so that it can be automatically corrected.
i was given a question without starting if its odd or even parity. the question is the decoder receives the code word r=1101101. determine whether an error has occurred and if so correct it. thats the question
At 9.43 to 9.52 seconds yur confusing with transmitter & receiver as send sending from receiver to transmitter so clear about ...other than this everything was perfect ...give clarity about that
Great lectures .....I become a fan of your lectures....but this lecture disappoint me a little ..why you didn't explain that how to choose data to find out P1,P2,P4 parity bit value.... it simply like to find P1.... start from position 1 and go on like use 1 and skip 1 bit...and so on..... However...all lectures are great sir.....Thanks again
@@srinityapadma5125 I don't remember exactly it's been a year and I have already passed the subject but it was something like both ways are correct both are just different representations of the same thing.
Question of most people->this doesnt make any sense. why does P1 depend on D3 D5 and D7, you arent explaining this at all. ANSWER The answer actually lies in the binary value of these positions lets say the value of position one is 1 in decimal and 001 in binary (considering that 7 can be represented in 3 bits), now Parity P1 lies on 001, we will select those values for which the 1 lies on this position, i-e the rightmost position (it may have more ones but 1 should be presented on the indicated position of the parity bit), naturally we will find these are D3 (011),D5(101), D7(111) so these three have one at the right most position. That is why we select them. Also the parity bits will have only Single 1 in them so issue of multiple ones will not be there, Practice it Hope it answers the question.
Thank you. So actually Mr Hamming was splitting the code to even and odd packet numbers and expects even and odd to the other side? And what if the parity code had noise and was transmitted faulty ?
Dislike because you didn't tell the logic behind how we did choose which parity bit is responsible of which data bits. Here is the logic behind: Parity bit P(2^n) checks data bits of which the nth rightmost bit has the value 1 (indexing starts from 0). Example: P4 checks 5(0101), 6(0110), 7(0111)
How data bit ia transmitted from receiver to Transmitter? I think the the correct way is data bit is transmitted from the Transmitter to receiver. And the receiver will do the correction.
+krishna Km Yeah that's what. I'm so confused at the moment. My teacher taught us in reverse order but different method. Practically, both answers should match regardless of the method but thats not the case :/ +Neso Academy help!
what happens if the parity bit itself picks up noise in the transmission process???if parity is checking the data error,how are we going to check the parity bit error if there's one???
Have you used the same yellow color as your green color? Or was it the same green color as yellow? Like, yes I can see the difference in those, but red and green would be better. Or heck, green and purple, anything, not this pair of same colors
why P1=D3 D5 D7 ?
Solutiion:
Plot 7 bits in binary & check in P1 row where is 1 ? its on LSB. So now check that entire LSB column where you find 1 and consider that data bit. ( follow " * " )
P1 0 0 1*
P2 0 1 0
D3 0 1 1 *
P4 1 0 0
D5 1 0 1*
D6 1 1 0
D7 1 1 1*
Hence, P1 = D3 D5 D7 and lly for P2, P3.
Hope helps you... :)
SAURABH BHAIJE grt thnk u
:)
Saurabh Bhaije thanks bro
i have that doubt only.that is clarified by u
I think this is not appy for p2 and p4
Neso Academy and Khan academy is the best out there
Is it so🙄
No it is not so.You are missing something bro.
😂yes bro
Now go and learn english from there
@@maniprakashv5213 ⁹⁹
Note: The associated data bits for the parity bit are chosen by the parity bit's exponentiation.
F.e. P1 = 2^0 = 1, so you check & skip bits always with 1 step. CHECK, SKIP, CHECK, SKIP etc. All your check-bits are the ones that will be used for parity later.
F.e. P2 = 2^1 = 2, this would result in CHECK, CHECK, SKIP, SKIP, CHECK, CHECK, SKIP, SKIP etc.
F.e. P3 = 2^2 = 4, this would result in 4 times CHECK, 4 times SKIP, 4 times CHECK, 4 times SKIP etc.
And the bits that are checked always have to be data bits. Try it with the 7 bit array in the upper right corner of the video. You will see how the associations
P1 -> D3,D5,D7
P2 -> D3D6D7
P4-> D5D6D7 are easily made.
firt we will check or skip i have dought how u calculate p1 p2 p4
good logic, easy to understand
Thanks
thanks a ton
Thanks broo
Position 1:
check 1 bit, skip 1 bit, check 1 bit, skip 1 bit, etc. (1,3,5,7,9,11,13,15,...)
Position 2 :
check 2 bits, skip 2 bits, check 2 bits, skip 2 bits, etc(2,3,6,7,10,11,14,15,...)
Position 4 :
check 4 bits, skip 4 bits, check 4 bits, skip 4 bits, etc(4,5,6,7,12,13,14,15....)
Position 8: check 8 bits, skip 8 bits, check 8 bits, skip 8 bits, etc. (8-15,24-31,40-47,...)
Thanks
Other way is to check and group similar bit group positions .
P1 index is 1 -> 001
Check of the bits with 1 at that position and group them ..
Similarly P2 is 010
Group again..
thank you!
Thankyou so much
your solution is the best. thx a lot bro
We don't have to cram anything out. p1 takes care of data bits which have 1 in rightmost place(i.e lsb) and p2 takes care of data bits which have 1 at second rightmost place after lsb and so on .Also for every r>=2,we have 2^r-1 length code block and length of message data is 2^r-r-1.
Thank God ...the only logical solution here.....
Thanks sir!! It's really helpful.. I think our hardware teacher must watch this video before giving that boring lecture to us !! anyway free of worry after watching your video!! Thanks Once again sir!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
U are correct
U were really studying 😂😂
SIR NOT ONLY YOUR VIDEOS BUT ALSO COMMENTS UNDER THEM MAKE US LEARN
superb, It Helped a lot for my exams.THANK U
The Hamming Code is an error-correcting code used to detect and correct single-bit errors in digital data transmission. To understand the process of using the Hamming Code for error detection and decision-making, let's break it down into stages:
1. Encoding: In this stage, the original data (message) is encoded using the Hamming Code. The encoding process adds extra parity bits (check bits) to the original data. These parity bits are strategically placed to help detect and correct single-bit errors. The total number of bits in the encoded message will be greater than the original data.
2. Transmission: After encoding, the message is transmitted through a communication channel. During transmission, there is a possibility of errors, such as flipping a single bit, due to noise or other factors.
3. Receiving: The encoded message is received at the destination. However, due to the transmission errors, the received message may have one or more bits in error.
4. Error Detection: In this stage, the received message is checked for errors using the parity bits added during encoding. The receiver calculates the parity of the received message and compares it with the parity bits. If the parity bits match the calculated parity, it indicates that no errors occurred during transmission. If the parity bits do not match, it means that there is at least one error in the received message.
5. Error Correction: If an error is detected in the previous step, the receiver uses the information from the parity bits to identify the location of the error and correct it. The Hamming Code allows for the correction of single-bit errors.
6. Decision-making: After error correction (if applicable), the receiver decides whether the message is error-free and ready for use or if it needs further action, such as retransmission of the message.
In summary, the stages involved in error detection and decision-making using the Hamming Code include encoding, transmission, receiving, error detection, error correction, and decision-making based on the corrected or uncorrected message.
hey you are so gooddddd at teaching .Thanks for saving my mid term test... Imagine spending 5 weeks listening to lecture but you cant learn shit.But a 12 minutes video make you undesratnd whole concept :D
this doesnt make any sense. why does P1 depend on D3 D5 and D7, you arent explaining this at all
+Andrew Morris the answer actually lies in the binary value of these positions lets say the value of position one is 1 in decimal and 001 in binary (considering that 7 can be represented in 3 bits), now Parity P1 lies on 001, we will select those values for which the 1 lies on this position, i-e the rightmost position (it may have more ones but 1 should be presented on the indicated position of the parity bit), naturally we will find these are D3 (011),D5(101), D7(111) so these three have one at the right most position. That is why we select them. Also the parity bits will have only Single 1 in them so issue of multiple ones will not be there, Practice it Hope it answers the question.
+Ussama Zafar Now THIS is what I needed! Thank you!
Ussama Zafar what if I want to calculate P5?
Gayatri Kulkarni First question is when will P5 needed? When a longer bits integer is used right? so then longer or bigger number will be represented in more binary digits i-e more then 3 bits (000,010 etc) so then binary digits will increase to the minimum no of binary digits to represent that number and that will how you will represent P5
Ussama Zafar gotcha ..thank you so much for your help 😃 ✌
Nice Approach Please please explain the scenario what will happen if the parity bit is changed?
did you get the ans to this?
When they uploaded this video, I was in sixth standard and now I am Btech first year student seeing their video.
very clear explanation..thanks a lot Neso Academy, your lectures are great !!!!!
Itne acche video ko dislike mat karo..bcz isko samajh ne ke liye toh thodi khudki v common sense chahiye😊
Really Nice explanation. Understood better than college professor.
Sir your intro music is so awesome
ur videos are unique and easy to understand
do u have have any videos on circular , bch , linear codes ?
pl reply
btw ur videos are amazing
Finally someone that will take criticism and help save their failing restaurant
Should the numbering be from right to left always? If we numering from left to right, we get another results
finally, I found someone who has the same problem :)
@@حاتمأبوحمّور fr
So does that mean he did mistake in the example taken right?
Great Lecture..explained so easily!!
what is the general rule of p1 ,p2 etc dependance on D respectively
Why P1 is associated with D3D5D7 only
thank you! You are very clear! Is there any rule to determine which data bits are controlled by parity bits?
Let a data bit sequence M=1110100001 is transmitted but the receiver receives the sequence with any
one bit corrupted. Use hamming code to identify the corrupted bit position so that it can be automatically
corrected.
Best explanation
bhai ek min or lamb karke P1,P2,P3 ka determination Bhi bata data, appreciate your work tho
such a wonderful explanation. I am greatly benefited.
sir please upload MEMORY DEVICES(rom, prom, ram &pram) lecture.
i was given a question without starting if its odd or even parity. the question is the decoder receives the code word r=1101101. determine whether an error has occurred and if so correct it. thats the question
Transmitter should be the one that transmits the data and Receiver should only receive.
At 9.43 to 9.52 seconds yur confusing with transmitter & receiver as send sending from receiver to transmitter so clear about ...other than this everything was perfect ...give clarity about that
good explanation !
Great lectures .....I become a fan of your lectures....but this lecture disappoint me a little ..why you didn't explain that how to choose data to find out P1,P2,P4 parity bit value.... it simply like to find P1.... start from position 1 and go on like use 1 and skip 1 bit...and so on.....
However...all lectures are great sir.....Thanks again
What if we right from left to right.
P1 P2 D3 P4 D5 D6 D7
Many books explain like this. Which one is correct?
Did you get the answer to this?
@@srinityapadma5125 Yes
@@prateekgupta2864 What is the answer?
@@srinityapadma5125 I don't remember exactly it's been a year and I have already passed the subject but it was something like both ways are correct both are just different representations of the same thing.
@@prateekgupta2864 Oo kk Fine Thanq
Thank for your videos
FYI, receiver DOESN'T transmit the data bits... other than that, good lecture... (y)
chutiye correction dikhta nhi kya
i love Neso Academy Videos of Digital electronics as well as of networking concepts :*
Thanks bro.... simple and informative 🙂
Plot of the binary level code issue from the homming code and error code is the expansion between the homming code
Sir board you will use while explaining the concepts
Superb video ..fully explained..
what if the transmitter transmitted a bad parity bit but the data bits are correct? Isn't this a flaw to the algorithm?
How to check the correctness of data if parity bit itsleves changes in data transmission?
Question of most people->this doesnt make any sense. why does P1 depend on D3 D5 and D7, you arent explaining this at all.
ANSWER
The answer actually lies in the binary value of these positions lets say the value of position one is 1 in decimal and 001 in binary (considering that 7 can be represented in 3 bits), now Parity P1 lies on 001, we will select those values for which the 1 lies on this position, i-e the rightmost position (it may have more ones but 1 should be presented on the indicated position of the parity bit), naturally we will find these are D3 (011),D5(101), D7(111) so these three have one at the right most position. That is why we select them. Also the parity bits will have only Single 1 in them so issue of multiple ones will not be there, Practice it Hope it answers the question.
Beautifully delivered!
thank u for good explanation .
Thank you. So actually Mr Hamming was splitting the code to even and odd packet numbers and expects even and odd to the other side? And what if the parity code had noise and was transmitted faulty ?
Did u got the answer. 'coz i have the same doubt
this is incomplete explanation what if parity bit changed due to noise? correct data will be interpreted wrong?
Same doubt here
Can noise cause any change to the parity bits?
your explanation is good! so please help me in keeping 15 bit hamming code generator circuit diagram and explanation :)
how parity bits value depends on D3 D5 D7?
can i know what is advantages n disadvantages of hamming code???
Excellent work Brother.
I managed to understand it from you in 13 mins and not from my uni teachers in a time span of 3 hours.😂
Dislike because you didn't tell the logic behind how we did choose which parity bit is responsible of which data bits.
Here is the logic behind:
Parity bit P(2^n) checks data bits of which the nth rightmost bit has the value 1 (indexing starts from 0).
Example: P4 checks 5(0101), 6(0110), 7(0111)
for better understand, show this ua-cam.com/video/vYGO5GU5A7o/v-deo.html
wonderfl explntion in 4min
How data bit ia transmitted from receiver to Transmitter? I think the the correct way is data bit is transmitted from the Transmitter to receiver. And the receiver will do the correction.
Thank you so much sir
my confusion are cleared thankyou :-)
Q: - 00111101010 What are the values of Parity Bits?
Like in video you said: P1=D3, D5, D7
What will happen if more than one bit will change???
first part of this tutorial missing
How we will find the errors if data bit and parity bit both will change..??How we can identify them??
my doubt is that if parity bit p1 and p2 changes such that the final parity remains same then how are we going to detect error?
Same doubt after 4 years😂
very very helpful and simple 👌 thank you 😇
how to know P1 is dependent on D3 D5 D7 WHY NOT D6. PLEASE EXPLAIN
Thnk u so much sir..😊😊😊
You dont explain how the p1, p2 and p4 chose to operate. How can one possibly understand that without explanation?
Dear presenter kindly check you use the word receiver at the place of sender??
how can the receiver send any signal........as u said in this video but in the previous one u said tansmitter send the signal
thanks for the hands-on
thank you sir
Sir why taking p1=D3, D5, D7.....can u plg explain reason
Thank u sir
Please explain clearly....
Thank you ...
And what if the noise is added to the parity bit itself? Then will it not result in unwanted modification in the original signal?
🎉 thankyou so much sir ❤
how is P1 related to d3 ,d5 and d7 ?
Happy teacher's day
sir how to check that parity pit depends on which bits?? eg: for 11 bit codes what should be the dependency of parity and data bits?
Thank you sir...
Is there a chance of addition of noise to parity bit?
In 2022 who's here
Me
The sequence you are using here is wrong , and so is your answer for the example .
How one can exactly know about Which bit has error? Rather than the group of bits??
SIR,i think you are marking the position in reverse order..... 4 3 2 1
it should be 1234
+krishna Km Yeah that's what. I'm so confused at the moment. My teacher taught us in reverse order but different method. Practically, both answers should match regardless of the method but thats not the case :/ +Neso Academy help!
I'm really confused :(
Shikhar Guptas
Same here. I go to UTA and my professor taught the reverse order. This guy is wrong
Ha mujhe bhi confuse kr diya tha
what happens if the parity bit itself picks up noise in the transmission process???if parity is checking the data error,how are we going to check the parity bit error if there's one???
It doesn't matter, the algorithm works just fine, see 3blue1brown video for explanation.
thank you very much
Thanks Sir😃😃
You have a funny accent but still you helped me a lot! Thanks! :)
Keep doing that.
How to find out Value of
P1 =D3 D5 D7
P2=D3 D6 D7
P4 =D5 D6 D7
thank you! this helped me so much!
seme
best video on this topic!! thanks sir!
Tnq sir
Such a great lecture!!!
the only slide i couldnt understand
Do you have lecture of DIJKSTRA's algorithm?
Have you used the same yellow color as your green color? Or was it the same green color as yellow? Like, yes I can see the difference in those, but red and green would be better. Or heck, green and purple, anything, not this pair of same colors
Very helpful. May the Lord Jesus Christ bless you.
what if there is an error in the parity bit?